A) - 1
B) 0
C) 1
D) 2
Correct Answer: A
Solution :
\[{{\left[ \frac{1+\cos (\pi /8)+i\sin (\pi /8)}{1+\cos (\pi /8)-i\sin (\pi /8)} \right]}^{8}}\] \[={{\left[ \frac{2{{\cos }^{2}}\left( \pi /16 \right)+2i\sin \left( \pi /16 \right)\cos \left( \pi /16 \right)}{2{{\cos }^{2}}\left( \pi /16 \right)-2i\sin \left( \pi /16 \right)\cos \left( \pi /16 \right)} \right]}^{8}}\] \[=\frac{{{[\cos \left( \pi /16 \right)+i\sin \left( \pi /16 \right)]}^{8}}}{{{[\cos \left( \pi /16 \right)-i\sin \left( \pi /16 \right)]}^{8}}}\] \[={{\left[ \cos \frac{\pi }{16}+i\sin \frac{\pi }{16} \right]}^{8}}{{\left[ \cos \frac{\pi }{16}+i\sin \frac{\pi }{16} \right]}^{8}}\] \[={{[\cos (\pi /16)+i\sin (\pi /16)]}^{16}}\] \[=\cos 16\left( \frac{\pi }{16} \right)+i\sin 16\left( \frac{\pi }{16} \right)\] = \[\cos \pi =-1\].You need to login to perform this action.
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